If $E$ has non-integral $j$-invariant, under what conditions will there exist an isogenous curve with integral $j$-invariant?

Here, saying an element $a \in K$ is integral means that $a$ belongs to the ring of integers of $K$.

I feel that the following result may be relevant.

Let $E$ be an elliptic curve over a local field $K$. We say $E$ has potential good reduction over $K$ if there is a finite extension $K^{\prime}/K$ such that $E$ has good reduction over $K^{\prime}$. The $j$-invariant of $E$ is integral if and only if $E$ has potential good reduction over $K$.

2 Answers
2

As Kevin says, you can use Neron-Ogg-Shafarevich. Here's an alternative proof.

We can assume that the local field $K$ is algebraically closed and complete for the valuation $v$. Then $E$ is $v$-adically analytically isomorphic to a Tate curve, which means that there is a $v$-adic analytic isomorphism $E(K)\to K^*/q^{\mathbb{Z}}$, where $q\in K^*$ satisfies $|q|_v<1$. Further, $|j(E)|_v = |q|_v^{-1}$. Let $E'$ be isogenous to $E$, say $\phi:E\to E'$. In the analytic model, the kernel of $\phi$ is generated by two elements $\zeta$ and $Q$, where $\zeta$ is an $n$'th root of unity and $Q$ is some root of $q$, say $Q^m=q$. Then$$E'(K) \cong K^*/\zeta^{\mathbb{Z}}Q^{\mathbb{Z}} \cong K^*/Q^{n\mathbb{Z}},$$ where the second isomorphism is raising to the $n$'th power. Thus $E'$ has a Tate parametrization, and the analytic formula gives $$|j(E')|_v = |Q^n|_v^{-1} = |q|^{-n/m}_v > 1.$$ Hence $E'$ has non-integral $j$-invariant.

You can find a discussion of Neron-Ogg-Shafarevich in my Arithmetic of Elliptic Curves, Chapter VII, Section 7, but there's one step (the hardest step) of the proof that is not done there. That step is in my Advanced Topics in the Arithmetic of Elliptic Curves, as is a detailed discussion of Tate models in Chapter V.

By the criterion of Neron-Ogg-Shafarevich, an elliptic curve has good reduction if and only if its $\ell$-adic Tate module is unramified for $\ell\neq p$, where $p$ is the residue characteristic of $K$. This is clearly an isogeny invariant. Using the result you mentioned, it follows that integrality of the $j$-invariant is an isogeny invariant.